" Value of "(ln x)/(1+m-2n)=(ln y)/(m+n-21)=(ln z)/(n+1-2m)" then "xyz=

(log x)/(l+m-2n)=(log y)/(m+n-2l)=(log z)/(n+l-2m)

If (log x)/(y-z) = (log y)/(z-x) = (log z)/(x-y) , then prove that xyz = 1 .

If ("log"x)/(y - z) = ("log" y)/(z - x) = ("log" z)/(x - y) , then prove that xyz = 1.

The value of log_(x^(n))y^(m) is

Suppose x,y,z>1 then least value of log(xyz)[(log x)/(log y log z)+(log y)/(log x log z)+(log z)/(log x log y)]

The value of lim_(x rarr0^(+))x^(m)(log x)^(n),m,n in N is

I_(m, n)= int_(0)^(1) x^(m) (ln x)^(n) dx=

If f(x) = log ((m(x))/(n(x))), m(1) = n(1) = 1 and m'(1) = n'(1) = 2, " then" f'(1) is equal to